Show that $xf \in L^2(\mathbb{R})$, $f \in L^2(\mathbb{R})$ implies $f \in L^1(\mathbb{R})$ How can I show that $xf \in L^2(\mathbb{R})$, $f \in L^2(\mathbb{R})$ implies $f \in L^1(\mathbb{R})$? I was thinking about expanding some sequence that converges to $f$ but am not sure this is the most elegant solution.
 A: $$\int|f| = \int_{(-\infty, -1] \cup [1,\infty)} |f| + \int_{(-1,1)} |f|$$ 
Applying Holder's inequality, we have:
$$\int_{(-\infty, -1] \cup [1,\infty)} |f| = \int_{(-\infty, -1] \cup [1,\infty)} \frac1{|x|} |x||f| \le \sqrt{\int_{(-\infty, -1] \cup [1,\infty)} \frac1{x^2}} \sqrt{\int_{(-\infty, -1] \cup [1,\infty)} x^2|f|^2} < \infty$$
and
$$\int_{(-1,1)}|f| = \int_{(-1,1)}1 \cdot |f| \le \sqrt{\int_{(-1,1)} 1} \sqrt{\int_{(-1,1)} |f|^2} \le \sqrt 2 \|f\|_{L^2} < \infty$$
Thus, $f\in L^1$.
A: Apply the Cauchy-Schwarz inequality to the interval $[\epsilon,r]$, for $r>\epsilon > 0$.  We have:
$$
\left(\int_{\epsilon}^r|f(x)|\,dx\right)^2 = \left(\int_{\epsilon}^r \frac 1x \cdot |x\,f(x)|\,dx\right)^2  \leq \left(\int_\epsilon ^r \frac 1{x^2}\,dx\right)\left( \int_\epsilon^r |x\,f(x)|^2 dx\right)
$$
Some messing around with this idea will give you the desired upper bound inequality.
In particular: you can use the above to show that $f$ is integrable over $(-\infty,-1) \cup (1,\infty)$.  Then, since $|f|^2$ is integrable over $[-1,1]$, you can conclude that $|f|$ is integrable over $[-1,1]$.  All together, $|f|$ is integrable over $\Bbb R$, as desired.
